Problem 6

If $c$ is a real number and the negative of one of the solutions of $x^2-3x+c=0$ is a solution of $x^2+3x-c=0$ , then the solutions of $x^2-3x+c=0$ are $\textbf{(A) }1,~2\qquad \textbf{(B) }-1,~-2\qquad \textbf{(C) }0,~3\qquad \textbf{(D) }0,~-3\qquad \textbf{(E) }\frac{3}{2},~\frac{3}{2}$

Solution

We let the roots of the first equation be $r,s$ and the roots of the second equation be $s, -t$ . By Vieta's Formulas, $r+s=3$ and $s-t=-3$ , $rs=c$ and $-st=c$ . So, $r=t$ . Thus, $t+s=3$ , $s-t=3$ , so $t=0$ , and $s=3\Rightarrow \textbf{(C)}$ .~MathJams

1976 AHSME (Problems • Answer Key • Resources)
Preceded by 1975 AHSME	Followed by 1977 AHSME
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME Problems and Solutions

Page

Toolbox

Search

1976 AHSME Problems/Problem 6

Problem 6

Solution